python - Convert Latitude and Longitude to point in 3D space

Question

I need to convert latitude and longitude values to a point in the 3-dimensional space. I've been trying this for about 2 hours now, but I do not get the correct results.

The Equirectangular coordinates come from openflights.org. I've tried several combinations of cos and sin, but the result did never look like our little beloved earth.

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In the following, you can see the result of applying the conversion Wikipedia suggests. I think one can guess from context what c4d.Vector is.

def llarToWorld(latit, longit, altid, rad):
    x = math.sin(longit) * math.cos(latit)
    z = math.sin(longit) * math.sin(latit)
    y = math.cos(longit)
    v = c4d.Vector(x, y, z)
    v = v * altid + v * rad
    return v

Red: X, Green: Y, Blue: Z

One can indeed identify North- and South America, especially the land around the Gulf of Mexico. However, it looks somewhat squished and kind of in the wrong place..

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As the result looks somewhat rotated, I think, I tried swapping latitude and longitude. But that result is somewhat awkward.

def llarToWorld(latit, longit, altid, rad):
    temp = latit
    latit = longit
    longit = temp
    x = math.sin(longit) * math.cos(latit)
    z = math.sin(longit) * math.sin(latit)
    y = math.cos(longit)
    v = c4d.Vector(x, y, z)
    v = v * altid + v * rad
    return v

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This is what the result looks like without converting the values.

def llarToWorld(latit, longit, altid, rad):
    return c4d.Vector(math.degrees(latit), math.degrees(longit), altid)

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Question: How can I convert the longitude and latitude correctly?

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Solution

Thanks to TreyA, I found this page on mathworks.com. The code that does it's work is the following:

def llarToWorld(lat, lon, alt, rad):
    # see: http://www.mathworks.de/help/toolbox/aeroblks/llatoecefposition.html
    f  = 0                              # flattening
    ls = atan((1 - f)**2 * tan(lat))    # lambda

    x = rad * cos(ls) * cos(lon) + alt * cos(lat) * cos(lon)
    y = rad * cos(ls) * sin(lon) + alt * cos(lat) * sin(lon)
    z = rad * sin(ls) + alt * sin(lat)

    return c4d.Vector(x, y, z)

Actually, I switched y and z because the earth was rotated then, however, it works! That's the result:

Answer 1

I've reformatted the code that was previously mentioned here, but more importantly you have left out some of the equations mentioned in the link provided by Niklas R

def LLHtoECEF(lat, lon, alt):
    # see http://www.mathworks.de/help/toolbox/aeroblks/llatoecefposition.html

    rad = np.float64(6378137.0)        # Radius of the Earth (in meters)
    f = np.float64(1.0/298.257223563)  # Flattening factor WGS84 Model
    cosLat = np.cos(lat)
    sinLat = np.sin(lat)
    FF     = (1.0-f)**2
    C      = 1/np.sqrt(cosLat**2 + FF * sinLat**2)
    S      = C * FF

    x = (rad * C + alt)*cosLat * np.cos(lon)
    y = (rad * C + alt)*cosLat * np.sin(lon)
    z = (rad * S + alt)*sinLat

    return (x, y, z)

Comparison output: finding ECEF for Los Angeles, CA (34.0522, -118.40806, 0 elevation)
My code:
X = -2516715.36114 meters or -2516.715 km
Y = -4653003.08089 meters or -4653.003 km
Z = 3551245.35929 meters or 3551.245 km

Your Code:
X = -2514072.72181 meters or -2514.072 km
Y = -4648117.26458 meters or -4648.117 km
Z = 3571424.90261 meters or 3571.424 km

Although in your earth rotation environment your function will produce right geographic region for display, it will NOT give the right ECEF equivalent coordinates. As you can see some of the parameters vary by as much as 20 KM which is rather a large error.

Flattening factor, f depends on the model you assume for your conversion. Typical, model is WGS 84; however, there are other models.

Personally, I like to use this link to Naval Postgraduate School for sanity checks on my conversions.